$A$ parallel plate capacitor has a capacitance of $10 \ \mu F$ with air between its plates. Now,half of the space between the two plates is filled with a dielectric material of dielectric constant $K = 4$ as shown in the figure. Find the capacitance of the capacitor in $\mu F$.

$A$ parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $k(x)=K(1+\alpha x)$ where $x$ is the distance measured from one of the plates. If $(\alpha d) << 1$,the total capacitance of the system is best given by the expression:

As shown in the figure,a dielectric slab of thickness $d/2$ and dielectric constant $K$ is placed between the plates of a parallel plate capacitor of plate separation $d$. The capacitor is charged to a potential $V$ using a battery. If the dielectric slab is pulled out after disconnecting the battery from the capacitor,the final potential difference across the plates of the capacitor is

The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used,is $C$. If the dielectric is replaced by another with dielectric constant $20$,the capacity will become

The capacitance of a capacitor is $1\,pF$ when there is air between its plates. If the distance between the plates is doubled and the space between them is filled with wax,the new capacitance becomes $2\,pF$. What is the dielectric constant of the wax?

$A$ parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ and $\varepsilon_2$,as shown in the figures. The distance between the plates is $d$ and the area of each plate is $A$. If the capacitance in the first configuration and second configuration are $C_1$ and $C_2$ respectively,then $\frac{C_1}{C_2}$ is